Prove that the series, $\sum_{n=1}^{\infty}{x^{2n}\over n^2+x^{2n}}$ converges uniformly in $[-1,1]$

Consider the series $$\sum_{n=1}^{\infty}{x^{2n}\over n^2+x^{2n}}$$

I want to show that the series is uniformly convergent in $$[-1,1]$$.

Theorem: A series of functions $$\sum f_n$$ will converge uniformly on $$[a,b]$$ if there exist a convergent $$\sum M_n$$ of positive numbers such that for all $$x\in [a,b]$$ $$|f_n(x)|\le M_n$$ for all $$n$$.

Attempt:

For $$x\in [-1,1]\quad$$ $$|x^{2n}|\le 1\tag{1}$$.

Recall,

Reverse Triangle Inequality $$\begin{equation*} ||x|-|y||\le|x+y|. \end{equation*}$$

So $$n^2-|x^{2n}|\le|n^2+x^{2n}|$$ Therefore we have $$n^2-1\le |x^{2n}+n^2|\tag{2}$$

Using $$(1)$$ and $$(2)$$, we have $$|f_n(x)|\le {1\over n^2-1}$$

Let $$a_n=1/n^2$$, then $${M_n\over a_n}={n^2\over n^2-1}={1\over 1-{1\over n^2}}$$ Therefor by Limit form comparison test, $$\sum M_n$$ converges.

The result follows.

Is this series uniformly convergent only in $$[-1,1]$$ and to what function it converges?

Edits: For $$|x|>1$$,

Divide by $$|x|^{2n}$$, we get $${1\over 1+{n^2\over |x|^{2n}}}$$. Since exponential grows faster than $$n^2$$ the expression goes to $$1$$ as $$n\to \infty$$

Therefore series diverges for $$|x|>1$$

• Does the summand go to zero for $x>1$, as $n$ goes to infinity? – AnyAD Oct 23 '18 at 12:05
• @AnyAD Thanks I understand it now. I have edited the post to show the calculation. – StammeringMathematician Oct 23 '18 at 12:24

Define $$f_n(x)={x^{2n}\over n^2+x^{2n}}$$In the interval $$[-1,1]$$ we have $$0\le x^{2n}\le 1$$ therefore $$|f_n(x)|=f_n(x)\le {1\over n^2+1}$$Now define $$M_n={1\over n^2+1}$$. Obviously the series $$\sum M_n$$ is convergent because$$\sum_{n=1}^{\infty}M_n=\sum_{n=1}^{\infty}{1\over n^2+1}\le \sum_{n=1}^{\infty}{1\over n^2}={\pi^2\over 6}$$and $$|f_n(x)|\le M_n$$which means that the sequence of $$\{M_n\}$$ fulfills the conditions of the theorem and using that theorem, we have proved what we wanted.
• I think you want $f_n(x)\le \frac{1}{n^2}.$ – zhw. Nov 30 '18 at 16:55
• Also ${1\over n^2}$ is another possibility for $M_n$ since ${1\over n^2+1}<{1\over n^2}$ – Mostafa Ayaz Nov 30 '18 at 16:57
• But why do you say $$f_n(x)\le {1\over n^2+1}?$$ – zhw. Nov 30 '18 at 17:01
• I'm just looking for some $M_n$ so that the conditions of the theorem hold i.e. $$|f_n(x)|\le M_n$$since we have $$|f_n(x)|\le {1\over n^2}$$ we can arbitrarily choose $M_n={1\over n^2+1}$ – Mostafa Ayaz Nov 30 '18 at 17:05
• But $1/(n^2+1)<1/n^2.$ – zhw. Nov 30 '18 at 17:08